Hazem Raw Mix Composition

LSF

Raw Mix Composition

& Quality Control

1.1 Hydraulic Modulus

HM =

CaO

= 1.7 --- 2.3

SiO

+ AL

O

+ Fe

O

It was found that with increasing HM, more heat is required for clinker burning : the strengths , especially the initial strengths step up and also the heat of hydration rises and the resistance to chemical attack decreases.

1.2 Silica Ratio

SR =

SiO

= 1.9 --- 3.2

Al

O3 + Fe

O

It was found that with increasing SR impairs the burnability of the clinker, by reducing the liquid phase content and tendancy towards formation of coating in the kiln, also causes a slow setting and hardening of the cement with decreasing SR the content of liquid phase increases: this improves the burnability of clinker and formation of coating in the kiln.

1.3 Alumina Ratio

AR =

Al

O

= 1.5 --- 2.5

Fe

O

the Alumina Ratio determines the composition of liquid phase in the clinker a high AR together with a low silica ratio results among other things,

in a fast setting of cement: this requires the addition of higher gypsum rate to control the setting time.

1.4 Lime Saturation Factor

to obtain complete lime saturation factor in clinker, the total silica must be combined as C3S, all iron onide must combine with the equivalent amount of alumina to form C4AF, and the remaining alumina must combine to C3A. the lime saturation factor is based on the assumption that from the

sintering temperature the clinker cools down at such slow rate that during

However, this is not the case with clinkers containing C3A. At the sintering temperature af about 1450 C, the silicate minerals C3S and C2S and possibly not transposed free lime are in a solid state, whereas C3A and C4AF are in a state of fusion.

how ever, the liquid phase is lower in lime that it would result from participation in C3A to complete the C3A the lime deficiency can be restored by extracting the lacking lime during the crystallization process from the solid phases, which simultaneously are the completed during fast technical clinker cooling: partically, the liquid aluminate cannot bind more lime than it already has absorbed at sentering temperature,experimental investigation showed that the most lime saturated liquid aluminate binds practically two molecules CaO per Al2O3. Therefore, under technical conditions, this is attainable lime limit, that called "standarad lime"

CaO = 2.8 SiO2 + 1.1 Al2O3 + 0.7 Fe2O3

stand

the lime saturation factor is the ratio of the real lime content to the standard lime:

L.S.F =

100 CaO

2.8 SiO2 + 1.18 Al2O3 + 0.65 Fe2O3

The Lime Saturation Factor is also part of the British Standard Specification and serves for the definition of admissible lime content.

L.S.F =

CaO - 0.7 SO3

2.8 SiO2 + 1.2 Al2O3 + 0.65 Fe2O3

in this formula LSF refers to the finished cement. The factor 0.7 SO3 is the CaO - content which equals analytically estimated SO3 - content , should be substracted from the total CaO - content.

it is assumed that the total SO3 comes from the added Gypsum and not from the clinker.

a high lime standard normally causes high cement strength 90 --- 95.

2. Allegation Alternate Method.

the simplest course of calculation for solving blending problems. This method allows the determination of proportion of two raw materials

91 45

76

NO.1 NO.2 NO.3 NO.4

proportion of both components can be determined. Example : 1

what mixing proportion is required for limestone with 91% CaCO3 to get a raw mix with a CaCO3 content 76% . The procedure is as follows,

limestone

76 - 31 = 45 parts lime stone

91 - 76 = 15 parts clay to get a raw mix with CaCO3 contents of 76% , 45 parts of lime stone should be mixed with 15 parts of clay. Thus the proportion of the components in the raw mix

limestone : clay 45 : 15 or 3 : 1

3. Calculation based on the Hydraulic Moodulus.

hydraulic module selected for clinker. To simplify the following calculations symbols are used for the designations of the clinker components,

Table 1 CaO C Cm C1 C2 SiO2 S Sm S1 S2 S3 Am A1 A2 A3 C3 C4 S4 A4 Fe2O3 F Fm F1 F2 F3 F4 Al2O3 A

so:

C

=

C

( for clinker )

S + A + F

HM =

C

( for raw mix )

Sm + Am + Fm

since both moduli are of the same value, one can equate.

HM =

C

=

C

S + A + F

Sm + Am + Fm

according to this method of calculation, it is assumed that X parts of the first material apportioned to one part of the second raw material.

under this assumption, the quantities of the particular raw material can be calculated by using the following formulas :

C

C

+ C

S

S

+ S

X

+ 1

+ 1

A

A

+ A

F

F

+ F

X

+ 1

+ 1

inserting the values Cm , Sm , Am , Fm into the formula for Hydraulic Moduli, we get

HM =

C

+ C

+ 1

S

+ S

A

+ A

F

+ F

+ 1

+ 1

+ 1

since the oxide components are known from the chemical analysis of raw material,and the Hydraulic Module is selected according to the quality requirements,the only remaining unknown is X .after transformation of the above formula, to calculate the value for X ,the following formula appears:

X = HM ( S2+A2+F2 ) - C2 = C2 - HM (S2+A2+F2)

i.e 1 2 3 4 5 6 limestone marl 76.87% 23.13% 100% SiO2 8.70 33.01 6.69 7.64 14.33 21.94 Al2O3 2.35 7.31 1.81 1.69 3.50 5.36 Fe2O3 1.32 4.83 1.01 1.12 2.13 3.26 CaO 47.80 30.22 36.75 6.99 43.74 66.92 MgO 1.50 0.66 1.15 0.15 1.30 2.00 SO3 0.30 0.20 0.23 0.05 0.28 0.44 LOI 37.96 23.77 29.18 5.49 34.67 .. balance 0.01 .. 0.05 .. 0.05 0.08 total 100.00 100.00 100.00 23.13 100.00 100.00 Example 2

two raw materials are given with the following composition (see column 1 and 2 of table 2 ).

calculate the composition of the raw mix, assuming a hydraulic module of HM = 2.2

solution:

X = 2.2 ( 33.01 + 4.31 + 4.83 ) - 30.22 Ξ 3.324 47.80 - 2.2 (8.72 + 2.35 +1.32 )

to get a clinker with a HM = 2.2 we have to mix 3.324 parts lime stone with one part of marl, or expressing as % , thus the raw mix consists of : 76.87% limestone, and 23.13 % marl.

Table 2

comp

limestonemarl raw

mix clinker

in the table 2 the calculated raw mix components appear in column 3 and 4, and the composition of raw mix is given in column 5 (column 3 + 4 = column 5 ) column 6 contains the calculated clinker composition as raw mix of raw of column 5, free of loss on iginition. From column 6 we obtain the value for the hydraulic module HM = 2.2

1 2 4

Lst clay clinker

X

+ 1

+ 1

LSF =

100 CaO

2.8 SiO2 + 1.18 Al2O3 + 0.65 Fe2O3

given 2 raw materials (see table 3, column 1 and 2 ).

X

F

+ F

X

+ 1

+ 1

To obtain a raw mix with a given LSF of 0.92, let us assume that X part of lime stone will apportioned to 1 part of clay.

LSF for

C

C

+ C

S

S

+ S

A

A

+ A

F

After transformation of the above formula, to calculate the value of X, the

LSF =

100 XC1 + C2

X + 1

following formula appears :

X = 100C2 - LSF (2.8S2 + 1.18A2 + 0.65F2) LSF (2.8S1 + 1.18A1 + 0.65F1) - 100C1

accordingly we get

X = 100*1.4 - 0.92(2.8*62.95 + 1.18*18.98 + 0.65*7.37) Ξ 3.85 0.92(2.8*1.42 + 1.18*0.48 + 0.65 *0.38) -100*52.6

thus 3.85 parts of limestone are apportioned to one part of clay raw mix consists of :

limestone clay 79.40% 20.60% Table 3 components 3 raw mix

1.42 62.95 21.96 0.48 18.98 6.68 0.38 7.37 2.84 52.6 1.4 65.51 1.11 0.98 1.69 0.85 0.85 1.32 43.16 7.47 .. 100 100 100 Note :

S

A + F

clinker % = 100 × Raw mix

100 - LOI

5. Calculation based on Lime Saturation Factor and

Silica Ratio

Example 4

to calculate a raw mix composed of limestone, slag and pyrite. The required LSF is 0.92 and S.M is 2.5 . The analysis of raw materials are shown in table 4 let's assume that x parts of limestone will be apportioned to y parts and 1 part of pyrite.

L.S.F =

100 C

SM =

2.8 S + 1.2 A + 0.65 F

the values of C,S,A and F are inserted in the above expressions :

L.S.F =

2.8(xS1+yS2+S3)+1.18(xA1+yA2+A3) + 0.65 (xF1+yF2+F3)

SM =

(xA1+yA2+A3) + (xF1+yF2+F3)

Expliciting x and y in the above equations we get :

x [ LSF (2.8 S1 +1.18 A1 +0.65 F1) - 100C1 ] + y [ LSF (2.8 S2 +1.18 A2 +0.65 F2) - 100C2 ] Ξ 100C3 - LSF (2.8 S3 +1.18 A3 +0.65 F3)

c1 - a1x c2 - a2x

b1 b2

c1 - b1y c2 - b2y

a1 a2

Now applying the values of SiO2, Al2O3,Fe2O3 and CaO from table 4 we get : to simpify let's assume

a1 = LSF (2.8 S1 +1.18 A1 +0.65 F1) - C1 a2 = SM (A1 + F1 ) - S1

b1 = LSF (2.8 S2 +1.18 A2 +0.65 F2) - C2 b2 = SM ( A2 + F2 ) - S2

c1 = C3 - LSF (2.8 S3 +1.18 A3 +0.65 F3) c2 = S3 - SM (A3 + F3 )

then the system of two equations quated above will take the following form

a1x +b1y = C1 y = .= a2x +b2y = C2 x = .=

x

y

x

y

limestone slag pyrite

69.8154% 27.8597% 2.3249% Table 4 co nst itue nt 1 2 3 4 5 6 7 8 0.69815 0.27859 0.02324 L.stone slag pyrite

co nst

itue

SiO2 6.75 39.45 11.21 4.71 10.99 0.26 15.96 22.07 Al2O3 0.71 9.67 1.57 0.50 2.69 0.04 3.23 4.47 Fe2O3 1.47 0.67 83.72 1.03 0.19 1.95 3.17 4.38 CaO 49.80 42.09 0.87 34.77 11.73 0.02 46.52 64.33 MgO 1.48 7.36 0.64 1.03 2.05 0.01 3.09 4.27 SO3 0.10 0.70 1.36 0.07 0.20 0.03 0.30 0.42 LOI 39.65 0.00 0.63 27.68 0.00 0.01 27.69 0.00 rest 0.04 0.06 0.00 0.00 0.00 0.00 0.04 0.06 LSF 0.92 0.92 SM 2.50 2.50 total 100.00 100.00 100.00 69.79 27.85 2.32 100.00 100.00

the calculated figures of the above table prove the correctness of the method of calculation

6. Calculation based on Lime Saturation Factor ,

Silica Ratio and Alumina Modulus

to calculate a raw mix composed of four raw materials, the mechanism is the same applied to the previous problem.

x = parts of limestone , y = parts of clay , z = parts of slag or silica sand or bauxite, apportioned to 1 part of iron ore = w.

performing the calculations we get

a1 = LSF (2.8 S1 +1.18 A1 +0.65 F1) - C1 b1 = LSF (2.8 S2 +1.18 A2 +0.65 F2) - C2 c1 = LSF (2.8 S3 +1.18 A3 +0.65 F3) - C3 d1 = C4 - LSF (2.8 S4 +1.18 A4 +0.65 F4) a2 = SM ( A1 + F1 ) - S1 b2 = SM ( A2 + F2 ) - S2 c2 = SM ( A3 + F3 ) - S3 d2 = S4 - SM (A4 + F4 ) a3 = AM . F1 - A1 b3 = AM . F2 - A2 c3 = AM . F3 - A3 d3 = A4 - AM . F4

x =

y =

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

z =

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

w =

7. Calculation based on KH , Silica Ratio

and Alumina Modulus

KH =

2.8 SiO2

to calculate a raw mix composed of four raw materials, the mechanism is the same applied to the previous problem.

x = parts of limestone , y = parts of clay , z = parts of slag or silica sand or bauxite, apportioned to 1 part of iron ore = w.

performing the calculations we get

a1 = KH (2.8 S1 ) - ( C1 - 1.65 A1 - 0.35 F1 ) b1 = KH (2.8 S2 ) - ( C2 - 1.65 A2 - 0.35 F2 ) c1 = KH (2.8 S3 ) - ( C3 - 1.65 A3 - 0.35 F3 ) d1 = ( C4 - 1.65 A4 - 0.35 F4 ) - KH ( 2.8 S4 ) a2 = SM ( A1 + F1 ) - S1 b2 = SM ( A2 + F2 ) - S2 c2 = SM ( A3 + F3 ) - S3 d2 = S4 - SM (A4 + F4 ) a3 = AM . F1 - A1 b3 = AM . F2 - A2 c3 = AM . F3 - A3 d3 = A4 - AM . F4

x =

y =

w =

y =

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

z =